# Difference between revisions of "1985 AJHSME Problems/Problem 9"

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==Solution== | ==Solution== | ||

− | + | First doing the subtraction, we get <cmath>\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10}</cmath> | |

− | If you don't believe this, then write the | + | We notice a lot of terms cancel. In fact, every term in the numerator except for the <math>1</math> and every term in the denominator except for the <math>10</math> will cancel, so the answer is <math>\frac{1}{10}</math>, or <math>\boxed{\text{A}}</math> |

+ | |||

+ | If you don't believe this, then write rearrange the factors in the denominator to get <cmath>\frac{1}{10}\times\frac{2}{2}\times\frac{3}{3}\times\cdots\times\frac{9}{9}</cmath> | ||

Everything except for the first term is <math>1</math>, so the product is <math>\frac{1}{10}</math> | Everything except for the first term is <math>1</math>, so the product is <math>\frac{1}{10}</math> |

## Revision as of 22:52, 13 January 2009

## Problem

The product of the 9 factors

## Solution

First doing the subtraction, we get

We notice a lot of terms cancel. In fact, every term in the numerator except for the and every term in the denominator except for the will cancel, so the answer is , or

If you don't believe this, then write rearrange the factors in the denominator to get

Everything except for the first term is , so the product is